Question

Two sisters, sister A and sister B, play SCRABBLE with each other every evening. Sister A is a statistician, and she draws a random sample of 30 results from the 1,420 total games that have been played to construct a confidence interval estimate of p, the proportion of SCRABBLE games between her and her sister that she has won. Her 95% confidence interval estimate of p is LCL = 0.36, UCL = 0.69.
A 95% confidence interval estimate of the total number of games sister A has won out of the 1,450 games that have been played is LCL =___________ and UCL = ___________

The Illinois State Toll Highway Authority is conducting a study to estimate the proportion of low-income commuters who drive to work on a toll road. The project manager wants to estimate the proportion to within 0.03 with 95% confidence, and the project manager believes that p will turn out to be approximately 0.11.

A sample size no smaller than __________ is needed.

Answer 1

Answer:

First question: LCL = 522, UCL = 1000.5

Second question: A sample size no smaller than 418 is needed.

Step-by-step explanation:

First question:

Lower bound:

0.36 of 1450. So

0.36*1450 = 522

Upper bound:

0.69 of 1450. So

0.69*1450 = 1000.5

LCL = 522, UCL = 1000.5

Second question:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

The project manager believes that p will turn out to be approximately 0.11.

This means that $\pi = 0.11$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

The project manager wants to estimate the proportion to within 0.03

This means that the sample size needed is given by n, and n is found when M = 0.03. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.03 = 1.96\sqrt{\frac{0.11*0.89}{n}}$

$0.03\sqrt{n} = 1.96\sqrt{0.11*0.89}$

$\sqrt{n} = \frac{1.96\sqrt{0.11*0.89}}{0.03}$

$(\sqrt{n})^2 = (\frac{1.96\sqrt{0.11*0.89}}{0.03})^2$

$n = 417.9$

Rounding up

A sample size no smaller than 418 is needed.